An algebra problem by Wildan Bagus Wicaksono

Algebra Level 3

Let n ≥ 2017 n \ge 2017 and x x be natural numbers such that

3 n 2 − 3 ( n − 2 ) 2 ( 3 2 n − 2 + 1 ) ( 3 2 n − 2 − 1 ) n − 2 = 3 x \large \sqrt [ n-2 ]{ \frac { { 3 }^{ { n }^{ 2 } }-{ 3 }^{ { (n-2) }^{ 2 } } }{ \left( { 3 }^{ 2n-2 }+1 \right) { (3 }^{ 2n-2 }-1) } } ={ 3 }^{ x }

Determine the value of x − n x - n .


The answer is -2.

Wildan Bagus Wicaksono by Wildan Bagus Wicaksono , 18, Indonesia

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2 solutions

3 x = 3 n 2 − 3 ( n − 2 ) 2 ( 3 2 n − 2 + 1 ) ( 3 2 n − 2 − 1 ) n − 2 Raise both sides to the power of n − 2 3 x ( n − 2 ) = 3 n 2 − 3 ( n − 2 ) 2 ( 3 2 n − 2 + 1 ) ( 3 2 n − 2 − 1 ) = 3 n 2 − 3 n 2 − 4 n + 4 3 4 n − 4 − 1 = 3 n 2 ( 1 − 3 − 4 n + 4 ) 3 4 n − 4 − 1 Multiply up and down by 3 4 n − 4 = 3 n 2 ( 3 4 n − 4 − 1 ) 3 4 n − 4 ( 3 4 n − 4 − 1 ) = 3 n 2 3 4 n − 4 = 3 n 2 − 4 n + 4 = 3 ( n − 2 ) 2 \begin{aligned} 3^x & = \sqrt [n-2] {\frac {3^{n^2}-3^{(n-2)^2}}{\left(3^{2n-2}+1\right)\left(3^{2n-2}-1\right)}} & \small \color{#3D99F6} \text{Raise both sides to the power of }n-2 \\ 3^{x(n-2)} & = \frac {3^{n^2}-3^{(n-2)^2}}{\left(3^{2n-2}+1\right)\left(3^{2n-2}-1\right)} \\ & = \frac {3^{n^2}-3^{n^2-4n+4}}{3^{4n-4}-1} \\ & = \frac {3^{n^2}\left(1-3^{-4n+4}\right)}{3^{4n-4}-1} & \small \color{#3D99F6} \text{Multiply up and down by }3^{4n-4} \\ & = \frac {3^{n^2}\left(3^{4n-4}-1\right)}{3^{4n-4}\left(3^{4n-4}-1\right)} \\ & = \frac {3^{n^2}}{3^{4n-4}} \\ & = 3^{n^2-4n+4} \\ & = 3^{(n-2)^2} \end{aligned}

Therefore, we have:

3 x ( n − 2 ) = 3 ( n − 2 ) 2 x = n − 2 ⟹ x − n = − 2 \begin{aligned} 3^{x(n-2)} & = 3^{(n-2)^2} \\ x & = n-2 \\ \implies x-n & = \boxed{-2} \end{aligned}

1 Helpful 0 Interesting 0 Brilliant 0 Confused

3 n 2 − 3 ( n − 2 ) 2 = 3 ( n − 2 ) 2 ( 3 n 2 − ( n − 2 ) 2 − 1 ) 3 n 2 − 3 ( n − 2 ) 2 = 3 ( n − 2 ) 2 ( 3 ( n + n − 2 ) ( n − n + 2 ) − 1 ) 3 n 2 − 3 ( n − 2 ) 2 = 3 ( n − 2 ) 2 ( 3 ( 2 n − 2 ) 2 − 1 ) 3 n 2 − 3 ( n − 2 ) 2 = 3 ( n − 2 ) 2 ( 3 4 ( n − 1 ) − 1 ) 3 n 2 − 3 ( n − 2 ) 2 = 3 ( n − 2 ) 2 ( 3 2 ( n − 1 ) + 1 ) ( 3 2 ( n − 1 ) − 1 ) { 3 }^{ { n }^{ 2 } }-{ 3 }^{ { (n-2) }^{ 2 } }={ 3 }^{ { (n-2) }^{ 2 } }\left( { 3 }^{ { n }^{ 2 }-{ (n-2) }^{ 2 } }-1 \right) \\ { 3 }^{ { n }^{ 2 } }-{ 3 }^{ { (n-2) }^{ 2 } }={ 3 }^{ { (n-2) }^{ 2 } }\left( { 3 }^{ (n+n-2)(n-n+2) }-1 \right) \\ { 3 }^{ { n }^{ 2 } }-{ 3 }^{ { (n-2) }^{ 2 } }={ 3 }^{ { (n-2) }^{ 2 } }\left( { 3 }^{ (2n-2)2 }-1 \right) \\ { 3 }^{ { n }^{ 2 } }-{ 3 }^{ { (n-2) }^{ 2 } }={ 3 }^{ { (n-2) }^{ 2 } }\left( { 3 }^{ 4(n-1) }-1 \right) \\ { 3 }^{ { n }^{ 2 } }-{ 3 }^{ { (n-2) }^{ 2 } }={ 3 }^{ { (n-2) }^{ 2 } }\left( { 3 }^{ 2(n-1) }+1 \right) \left( { 3 }^{ 2(n-1) }-1 \right)

3 x = 3 n 2 − 3 ( n − 2 ) 2 ( 3 2 n − 2 + 1 ) ( 3 2 n − 2 − 1 ) n − 2 3 x = 3 ( n − 2 ) 2 ( 3 2 ( n − 1 ) + 1 ) ( 3 2 ( n − 1 ) − 1 ) ( 3 2 n − 2 + 1 ) ( 3 2 n − 2 − 1 ) n − 2 3 x = 3 ( n − 2 ) 2 ( 3 2 ( n − 1 ) + 1 ) ( 3 2 ( n − 1 ) − 1 ) ( 3 2 ( n − 1 ) + 1 ) ( 3 2 ( n − 1 ) − 1 ) n − 2 3 x = 3 ( n − 2 ) 2 n − 2 3 x = 3 n − 2 x = n − 2 x − n = n − 2 − n x − n = − 2 \large { 3 }^{ x }=\sqrt [ n-2 ]{ \frac { { 3 }^{ { n }^{ 2 } }-{ 3 }^{ { (n-2) }^{ 2 } } }{ \left( { 3 }^{ 2n-2 }+1 \right) { (3 }^{ 2n-2 }-1) } } \\ \large { 3 }^{ x }=\sqrt [ n-2 ]{ \frac { { 3 }^{ { (n-2) }^{ 2 } }\left( { 3 }^{ 2(n-1) }+1 \right) \left( { 3 }^{ 2(n-1) }-1 \right) }{ \left( { 3 }^{ 2n-2 }+1 \right) { (3 }^{ 2n-2 }-1) } } \\ \large { 3 }^{ x }=\sqrt [ n-2 ]{ \frac { { 3 }^{ { (n-2) }^{ 2 } }\left( { 3 }^{ 2(n-1) }+1 \right) \left( { 3 }^{ 2(n-1) }-1 \right) }{ \left( { 3 }^{ 2(n-1) }+1 \right) { (3 }^{ 2(n-1) }-1) } } \\ \large { 3 }^{ x }=\sqrt [ n-2 ]{ { 3 }^{ { (n-2) }^{ 2 } } } \\ \large { 3 }^{ x }={ 3 }^{ n-2 }\\ \large x=n-2\\ \large x-n=n-2-n\\ \large x-n=-2

1 Helpful 0 Interesting 0 Brilliant 0 Confused

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